PHP populate drop box with jquery

Question

I have a script which fetches options from a script php to populate a drop down list on the main page.

Here's the javascript

   <script>
   //# this script uses jquery and ajax it is used to set the values in
           $(document).ready(function(){   
                //# the time field whenever a day is selected. 
                $("#day").change(function() {   

                      var day=$("#day").val();
                      var doctor=$("#doctor").val();

                      $.ajax({
                          type:"post",
                          url:"time.php",
                          data:"day="+day+"&doctor="+doctor,
                          dataType : 'json'
                          success: function(data) {
                                //# $("#time").html(data);
                                var option = '';
                                $.each(data.d, function(index, value) {
                                     option += '<option>' + value.timing + '</option>';
                                });
                                $('#timing').html(option);
                             }
                       });
                  });
             });
   </script>

Here's the php script which gets data from a database.

  <?php
    $con = mysqli_connect("localhost","clinic","myclinic","myclinic");

    // Check connection
    if (mysqli_connect_errno())
    {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $doctor = $_POST['doctor'];
    $day = $_POST['day'];

    $query = "SELECT * FROM schedule WHERE doctor='" .$doctor."'AND day='" .$day. "'";

    $result = mysqli_query($con, $query);

    //$res = array();

    echo "<select name='timing' id='timing'>";

    //Initialize the variable which passes over the array key values
    $i = 0;                                 

    //Fetches an associative array of the row
    $row = mysqli_fetch_assoc($result);

    // Fetches an array of keys for the row.    
    $index = array_keys($row);             

    while($row[$index[$i]] != NULL)
    {
        if($row[$index[$i]] == 1) {             
            //array_push($res, $index[$i]);
            json_encode($index[$i]);

            echo "<option value='"  . $index[$i]."'>" . $index[$i] . "</option>";
        }
        $i++;
    }       

    echo json_encode($res);

    echo "</select>";

  ?>

It's not working. I get an error from console saying missing '}' in javasrcipt on line

  $("#day").change(function(){

I can't seem to find an error either.

Answer 1

It's hard to say where is problem, because you mixed things together. On Javascript side you expect JSON but on PHP side you generate HTML.

Use JSON for sending data between server and browser. Ensure that you actually generate valid JSON and only JSON.

This line does nothing (function returns value, but not modifies it)

json_encode($index[$i]);

This line does not make sense - variable $res is not initialized;

echo json_encode($res);

Answer 2

You need to add a comma on the line above the one triggering the error :

dataType : 'json',

Answer 3

It's because you don't have a comma on the line above it...